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Find values of k if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2)

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#### Solution

Area of a triangle = `1/2 abs ((x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1))`

`=> 1/2 abs ((k,0,1),(4,0,1),(0,2,1)) = pm 4`

`=> abs ((k,0,1),(4,0,1),(0,2,1)) = pm 8`

`=> (-2) abs ((k, 1),(4,1)) = pm 8 ` ...[On expanding with respect toC_{2}]

`=> k - 4`

= `pm 4`

`=> "k" = 0, 8`

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